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Exercise 10.6 · Q8

Q.(x2+y2)dy=xy dx\left(x^2+y^2\right)dy=xy\,dx. It is given that y(1)=1y(1)=1 and y(x0)=ey(x_0)=e. Find the value of x0x_0.

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Solve the homogeneous equation for the general implicit solution, fix the constant using y(1)=1y(1)=1, then substitute y=ey=e into the particular solution and solve for x0x_0.

Step 1. Rewrite. dydx=xyx2+y2\dfrac{dy}{dx}=\dfrac{xy}{x^2+y^2} — homogeneous of degree 00.

Step 2. Substitute y=vxy=vx. v+xdvdx=v1+v2 ⟹ xdvdx=v−v(1+v2)1+v2=−v31+v2v+x\dfrac{dv}{dx}=\dfrac{v}{1+v^2}\ \Longrightarrow\ x\dfrac{dv}{dx}=\dfrac{v-v(1+v^2)}{1+v^2}=\dfrac{-v^3}{1+v^2}.

Step 3. Separate. 1+v2v3 dv=−dxx ⟹ (v−3+v−1)dv=−dxx\dfrac{1+v^2}{v^3}\,dv=-\dfrac{dx}{x}\ \Longrightarrow\ \left(v^{-3}+v^{-1}\right)dv=-\dfrac{dx}{x}.

Step 4. Integrate. −12v2+ln⁡∣v∣=−ln⁡∣x∣+C ⟹ −12v2+ln⁡∣vx∣=C-\dfrac{1}{2v^2}+\ln|v|=-\ln|x|+C\ \Longrightarrow\ -\dfrac{1}{2v^2}+\ln|vx|=C.

Step 5. Replace v=yxv=\dfrac{y}{x}: vx=y, v2=y2x2vx=y,\ v^2=\dfrac{y^2}{x^2}. −x22y2+ln⁡∣y∣=C-\dfrac{x^2}{2y^2}+\ln|y|=C. …

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